Hello together,

I have a rather basic issue on propositional logic: first, consider an arbitrary set of formulas $T$ that is consistent and complete, i.e., for every propositional formula $\varphi$, either there holds $T\vdash\varphi$ or $T\vdash\neg\varphi$.

My Question is: Is there a minimal consistent and complete subset $T^\prime\subseteq T$, i.e., $T^\prime$ is still complete, but no proper subset $T^{\prime\prime}$ of $T^\prime$ is?

I know some examples where that is the case, namely e.g. the set of propositional variables $T_1=\{A_0,A_1,\dots\}$. One can also vary this example by considering any subset $T_2$ of the literals (i.e., variables or negations of them) such that for any $i\geq0$, either $A_i\in T$ or $\neg A_i\in T$. In these cases, both $T_1$ and $T_2$ are already minimal in the above sense.

However, my attempts to prove the question above failed so far. My first approach was it to successively eliminate formulas $\varphi$ that can be proven from the rest, i.e. such that $T\setminus\{\varphi\}\vdash\varphi$. More detailed: let $T=\{\varphi_0,\varphi_1,\dots\}$. Then we define a sequence $(T_i)_{i\geq0}$ of subsets of formulas of $T$ as follows:

  1. $T_0:=T$

  2. If $T_i$ is defined, then search if there is an index $j\geq0$ such that $T_i\setminus\{\varphi_j\}\vdash\varphi_j$. If yes, pick the least such $j$ and define $T_{i+1}:=T_i\setminus\{\varphi_j\}$. Otherwise, set $T_{i+1}=T_i$

Now if the construction stops at some point, i.e. if there is an $i\geq0$ such that $T_i=T_{i+1}$, then for least such $i$, $T_i$ is minimal by construction. However, if the construction runs forever, then we cannot argue that the intersection $\bigcap_{i\geq0}T_i$ is still complete because the notion of proof is finite.

I have made now several other attempts, including infinite proofs and axiom of choice, but still nothing seemed to helped. Maybe I have overseen something. Still, I think that the answer to the question is Yes. Does somebody have an idea (or have a counterexample) for this question.

Yours sincerely,


  • $\begingroup$ It seems to me that you are assuming that $T$ is countable. $\endgroup$
    – user40023
    Feb 2, 2015 at 11:20
  • $\begingroup$ Yes, I consider only propositional formulas built up from a countable set of variables, i.e., w.l.o.g. let $\{A_0,A_1,\dots\}$ be the set of propositional variables. $\endgroup$ Feb 2, 2015 at 11:29
  • $\begingroup$ A related fact: every theory $T$ in classical propositional or first-order logic has an independent axiomatization, i.e., a set of formulas $T'$ such that $T$ and $T'$ generate the same theory, but no proper subset of $T'$ does. For countable $T$, the argument is quite simple: we enumerate $T=\{A_n:n<\omega\}$, and let $I$ be the set of $n$ such that $A_0,\dots,A_{n-1}\nvdash A_n$. Then one easily checks that $T'=\{A_0\land\dots\land A_{n-1}\to A_n:n\in I\}$ works. The uncountable case is a nontrivial theorem due to Reznikoff; see arxiv.org/abs/1108.5171 for an English translation. $\endgroup$ Feb 2, 2015 at 21:55

1 Answer 1


Let $\varphi_n=A_0\wedge A_1\wedge\cdots\wedge A_{n-1}$ be the assertion that the first $n$ many propositional variables are true. The theory $T=\{\varphi_n\mid n\in\mathbb{N}\}$ consisting of all these assertions is complete and consistent, but there is no minimal complete consistent subset of $T$, since $\varphi_k\to\varphi_n$ is a tautology when $n\leq k$, and so any particular formula can be omitted without loss from any infinite subcollection; but no finite set of the $\varphi_n$ is complete.

The same idea works even when there are uncountably many variables. Just divide them into groups of countably infinitely many and do the same trick within each group.


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